If $f:R \to R$ such that $(f(x))^7=x-f(x)$ find required area

Question

If $f:R \to R$ be a differentiable function such that $(f(x))^7=x-f(x)$, then find

find the area bounded by curve $y=f(x)$ between the ordinates $x=0$ and $x=\sqrt{3}$ and $x-$axis.

(A) $\frac{f(\sqrt3)}{8}[8 \sqrt{3}-(f(\sqrt3))^7-4 f(\sqrt3)]$

(B) $\frac{f(\sqrt3)}{8}[8 \sqrt{3}-(f(\sqrt3))^7]$

(C) $\sqrt3 f(\sqrt3)-\frac{93}{8}$

(D) None of these.

Here $(f(x))^7=x-f(x)$

Hence $x=(f(x))^7+f(x)$

which gives $f^{-1} (x)=x^7+x$

i.e. but there is no way to reach $f(x)$ itself. Could someone help me with this?

Answer 1

HINT

We know that $$\int_{a}^{b} f(x) dx+\int_{f(a)}^{f(b)} f^{-1}(x) dx =bf(b)-af(a)$$

This follows as substituting $f(x)=u$, we get $$\int_{a}^{b} f(x) dx+\int_{f(a)}^{f(b)} f^{-1}(x) dx = \int_{a}^{b} f(x) + \int _{a}^{b} u f'(u) du=\int_{a}^{b} (xf(x))' dx$$

As $(xf(x))'=f(x)+xf'(x)$ from product rule. So we get $$\int_{0}^{\sqrt{3}}f(x) dx+\int_{0}^{f (\sqrt{3})}x^7+x \; dx=\sqrt{3}f(\sqrt{3})$$ As $f(0)=0$ and the inverse of $f(x)$ is $x^7+x$. Can you take it from here?